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PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 426 PROBLEM 12.78 Determine the mass of the earth knowing that the mean radius of the moon’s orbit about the earth is 238,910 mi and that the moon requires 27.32 days to complete one full revolution about the earth. SOLUTION We have 2 [Eq. (12.28)]= Mm F G r and 2 n n v F F ma m r = = = Then 2 2 Mm v G m rr = or 2r M v G = Now 2 r v π τ = so that 2 2 32 1 2r r M r G G π π τ τ = = Noting that 627.32days 2.3604 10 sτ = = × and 9238,910 mi 1.26144 10 ftr = = × we have 2 9 3 9 4 4 6 1 2 (1.26144 10 ft) 34.4 10 ft /lb s 2.3604 10 s M π − = × × ⋅ × or 21 2413 10 lb s /ftM = × ⋅
